CodeForces 157B Trace

Sherlock Holmes发现了一面被几个同心圆分成多个区域的墙,这些区域交替涂成红色和蓝色。任务是计算所有红色区域的总面积。通过输入不同圆的半径并使用数学公式来解决这个问题。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

B. Trace
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.

Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.

Input

The first line contains the single integer n (1 ≤ n ≤ 100). The second line contains n space-separated integers ri (1 ≤ ri ≤ 1000) — the circles' radii. It is guaranteed that all circles are different.

Output

Print the single real number — total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.

Examples
input
1
1
output
3.1415926536
input
3
1 4 2
output

40.8407044967

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
double pi=2*asin(1.0);
int n;
double a[105];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lf",&a[i]);
        sort(a+1,a+n+1);
    double sum;
    if(n&1)
    {
        sum=pi*a[1]*a[1];
        for(int i=2;i<=n;i+=2)
        {
            sum+=pi*(a[i+1]*a[i+1]-a[i]*a[i]);
        }
    }
    else
    {
        sum=0;
        for(int i=2;i<=n;i+=2)
        {
            sum+=pi*(a[i]*a[i]-a[i-1]*a[i-1]);
        }
    }
    printf("%lf\n",sum);
    return 0;
}


转载于:https://www.cnblogs.com/dacc123/p/8228644.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值